According to principle of homogeneity dimension of LHS should be equal to dimensions of RHS so option (3) is correct. $\begin{aligned}
& \mathrm{T}^2=\frac{4 \pi^2 \mathrm{r}^3}{\mathrm{GM}} \
& {\left[\mathrm{T}^2\right]=\frac{\left[\mathrm{L}^3\right]}{\left[\mathrm{M}^{-1} \mathrm{L}^3 \mathrm{T}^{-2}\right][\mathrm{M}]}}
\end{aligned}(Dimensionof\mathrm{G}is\left[\mathrm{M}^{-1} \mathrm{L}^3 \mathrm{T}^{-2}\right])\left[\mathrm{T}^2\right]=\frac{\left[\mathrm{L}^3\right]}{\left[\mathrm{L}^3 \mathrm{~T}^{-2}\right]}=\left[\mathrm{T}^2\right]$
Applying the principle of homogeneity of dimensions, determine which one is correct, where T is time period, G is gravitational constant, M is mass,…
JEE Main 2024 — Physics Mechanics
2024mcqmedium
Applying the principle of homogeneity of dimensions, determine which one is correct, where T is time period, G is gravitational constant, M is mass, r is radius of orbit.
Official previous-year question
Held on 4 Apr 2024 · Verified 6 Jul 2026.
Options
- A
T2=GM4π2r2
- B
T2=GM24π2r
- C
T2=GM4π2r3
- D
T2=4π2r3
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